Question

Introductory chemistry

20 A nitric acid solution containing 71.0% HNO3 (by mass) has a density of 1.42 g/mL. ow many moles of HNO, are present in 2.36 L of this solution? b) Determine the molarity of the acid.

Answer 1

71g of HNO is present in 100 g of the Solution (a) Given volume = 2.36 h = 2.36X 1000 ml = 2360 ml - mass of the solution - volume x density = 2360ml x 1.42 g ma = 3351- 29 mass of HNO mass% of HNO x 100 mass of the solution mass of HNO3 - X 100 3351-2 mass of HNO = ?X 3351.2. loo mass of HNO₂ = 2379.3529 of 4 Noz

males of H Nez = mass Molar mass = 9379.3529 639/mot 32.767499 moler Molarity = moles of HNO2 Volume of HNO3 in L' 37.267492 holes 2.36L = 16 M.